examples sil

7/29/2019 Examples Sil

1/26

Line Models and SIL

1.0 Introduction

In these notes, I present different linemodels that are used, and I also make some

comments on Examples 4.2 and 4.3 leading

to discussion of surge impedance loading,

and finally I give a hint for problem 4.21.2.0 Simplified models (Section 4.5)

We recall two things. First, we have the so-called exact transmission line equations:

lIZlVV C sinhcosh 221 (1)

lZ

VlII

C

sinhcosh 221 (2)

Second, we may represent a transmission

line using a -equivalent model, shown

below, if we useZand Y/2, where

IZ

IY1 IY2

I1

V2

I2

Y/2

V1

Y/2

Z

Fig. 1

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l

lZlZZ C

sinhsinh' (3)

2/

)2/tanh(

2tanh2'

l

lYl

ZY

C

(4)

Note that the two are equivalent, i.e., use of

the -equivalent transmission model withZ

and Yis equivalent to using eqs. (1), (2).

Question: When is it OK to use the -

equivalent transmission model with Zand Y

(instead ofZand Y)?

(RecallZ=zl, Y=ylwhere lis line length).

Lets look at eqs. (3), (4) in more detail.

They tell us thatZZand YYwhen

1sinh

l

l(5)

12/

)2/tanh(

l

l(6)

2


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To see when this happens, lets look up in a

good math table how to express sinh(x) and

tanh(x) as a Taylors series. I used [

1

, p. 58-59] to find that:

!7!5!3sinh

753 xxxxx

4

315

17

15

2

3tanh

2

2

753

xxxx

xx

Using these in eqs. (5) and (6), we get:

1!7!5!3

753

l

llll

(7)

1

2/

315

2/17

15

2/2

3

2/2/

753

l

llll

(8)

For both of these equations, they become

true as the higher-order terms in the

numerator get small relative to the first term

in the numerator. This happens for small |l|,

which occurs for small line length l.

So when |l| is small, it is quite reasonable to

useZ=Z=zland Y=Y=yl.

3


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Consider a lossless line, i.e., a line for which

r=0 in z=r+jx and g=0 in y=g+jb. (Note

that for inductive series elements andcapacitive shunt elements, that x and b will

be positive numbers when defined with

positive signs inzandy). Then

xbgxrbjrgjbgjxrzy ++=++== )())((

For transmission lines, g=0 always. Soxbjrb = (9)

In the lossless case, r=0, and we get jxbjxb == (10)

For a lossless line (=j), it is possible to

show that =j=j0.0000013/meter [2, pg

211] is quite typical for most transmissionlines. For a 100 mile-long line:

2092.010016090000013.0

milesmile

meters

meterl

Then:2077.0)2092.0sinh( jj

105.0)1046.0tanh()2/2092.0tanh( jjj

which shows that

sinh(l)l

tanh(l/2) l/2

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as required.

The loss of accuracy from the approximation

in these cases can be seen from:9928.0

2092.0

2077.0sinh

l

l

0038.11046.0

105.0

2/

)2/tanh(

l

l

The text recommends that lines longer than

150 miles should use Long-Line model, but[2] recommends that lines longer than about

100 miles should use theLong-Line model.

Lines below about 100 miles may use Z=zl

and Y=yl. Doing so results in the Medium-Length model, sometimes also referred to as

the nominal -equivalent model.

A final model suggested by the text is theshort-length model, for lines shorter than 50

miles. This is the same as the Medium-Length model except Y is neglected

altogether. This makes sense from the point

of view that the parallel-plate capacitor in

5


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this case, can be considered to have short

length, and thus a small area of the plates.

3.0 Surge impedance loading

Recall our definition of characteristic

impedanceZC as:

y

zZC (11)

Example 4.2 considers a transmission line

terminated in its characteristic impedance,

per Fig. 2 (the long-line model is used).

y

zZC

IZ

IY1 IY2

I1

V2

I2

Y/2

V1

Y/2

Z

Fig. 2

One result of the analysis in Example 4.2 is

to show that the complex power gain (theratio of the power flowing out of the line to

the power flowing into the line) is given by:

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leP

P

S

S 2

12

21

12

21(12)

where is the attenuation constant(=+j).

This says that when a line is terminated in

ZC, the complex power gain (actually loss) is

purely real.

The implication of this is that the line (whenterminated inZC), only affects the real power

(decreases it) but does not affect the reactive

power at all.

Consider reactive power implication:Whatever reactive power flows out of the

line (and into the load) also flows into the

line. So a line terminated in ZC has a very

special character with respect to reactive

power: the amount of reactive power

consumed by the series X is exactlycompensated by the reactive power supplied

by the shunt Y, for every inch of the line!

7


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Another result of Example 4.2 is:

leV

V

1

2(13)

It is the case that is always non-negative.

This means that el>1 and 0


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So |V2|=|V1| for lossless line terminated in

ZC.

Note also that lossless implies

C

L

cl

ll

c

l

cj

lj

y

zZ indindindC (15)

SoZC is purely real for the lossless case.

So for lossless line terminated in ZC, since

ZC is purely real, then only real power is

delivered to it, and since the line is lossless,

this same real power is delivered from the

source. Therefore:

CC Z

V

Z

V

P

2

1

2

2

12 (16)The characteristic impedance is clearly an

important parameter. It is also commonly

referred to in the industry as the surge

impedance. And the power flowing into it,

per eq. (16), is called the surge impedanceloading, denoted by PSIL, i.e.,

C

SILZ

VP

2

1 (17)

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Equation (17) gives SIL in terms of per-

phase power (and line-to-neutral voltage).

The text also gives it in terms of 3-phasepower (and line-to-line voltage) as:

C

ll

SILZ

VP

2

13 (18)

Surge impedance and SIL are given for

typical overhead 60 Hz three-phasetransmission lines in Table 1 [3, 4].

Table 1

Vrated

(kV)

CLZC /

(ohms)

CratedSIL ZVP /2

(MW)

69 366-400 12-13

115 380* 35

138 366-405 47-52

161 380* 69

230 365-395 134-145

345 280-366 325-425

500 233-294 850-1075765 254-266 2200-2300

1100 231 5238

* Estimated

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4.0 Line limits (Section 4.9)

Fig. 3 is a well-known conceptual curve thatcaptures some attributes of transmission

lines (note that the numbers are typical and

limits for any particular line may vary).

Note the vertical axis is given as a

percentage of SIL, simply because SILprovides a convenient characteristic of a

transmission line that captures an attribute

related to its power handling capability as a

function of its physical construction.

However, SIL does not capture the influence

of length on power handling capability.

Fig. 3

11


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These attributes are:

Power limit decreases with line length

Short lines are limited mainly by thermalproblems.

Medium length lines tend to be limited by

voltage-related problems.

Very long lines tend to be limited by

stability problems.

5.0 Complex power expression

This material combines Sec. 4.6, 4.8, 4.9.

Consider the long transmission line of Fig.

4. The voltages at the ends are specified as:1

11

j

eVV2

22

j

eVV (19)The series impedance is

ZjeZZ (20)

LZ

IZ

IY1 IY2

I1

V2

I2

Y/2V1

Y/2

Z

Fig. 4

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The complex power transferred into the line

from bus 1 is:*2

1

*

112 2/YVIVS Z (21)But IZ can be expressed as:

Z

VVIZ

21 (22)

Substitution of (22) into (21) yields:

*2

1

*

21

112 2/YVZ

VVVS (23)

Distributing the V1 through, we obtain:

*2

1*

21

*

2

1

*2

1*

*

21

*

11

12

2/

2/

12

YVZ

eVV

Z

V

YVZ

VVVVS

j (24)

where 12= 1- 2. Eq (24) is the same as eq.

4.29, pg. 104 in the text except eq. (24) also

includes the effect of the sending-end line-

charging, as represented by the last term.

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6.0 Circle diagrams

If we ignore the last term in eq. (24), whichmeans assuming no charging capacitance,

and if we assume Z=Z, i.e., for the short-

line model, eq. (24) becomes:

*

21

*

2

1

12

12

Z

eVV

Z

VS

j

=

Using Z*=|Z|e-jZ, we get

12

||||

21

2

1

12

jZjZj eeZ

VVe

Z

VS = (24a)

We may repeat this same process to obtain

power from the other direction as

12

||||

21

2

2

21

jZjZj eeZ

VVe

Z

VS

=

The power flowing into the receiving bus

from the line is then just S21, or

12

||||

21

2

2

21

jZjZj eeZ

VVe

Z

VS

+= (24b)

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15/26

We can then define the following:

ZjZjZj e

Z

VVBe

Z

VCe

Z

VC ===

||||||

21

2

2

2

2

1

1

which are all constants if we assume the

voltages are fixed.

Then eqs. (24a) and (24b) become:12

112

jBeCS = (24c)

12

221

j

BeCS

+= (24d)Eqs. (24c) and (24d) characterize sending

end and receiving end complex power as a

function of the angle between the buses, 12.

C1 and C2 are the centers of circles having

radii |B|=|V1||V2|/|Z|.

Figure 4.7 illustrates circle diagrams. The

text makes 6 observations about these that I

encourage you to review, among which are:

#5: Transmission is strengthened by

increasing voltages (generator fieldwindings) and decreasing reactance (series

compensation)

15


16/26

#6: Strong coupling between P-flow and 12and between Q-flow and |V|.

7.0 An alternative developmentWe want simple expressions for P12 and Q12.

Recall eq. (3) above, repeated here:

l

lZlZZ C

sinhsinh' (3)

In the lossless case, =j, so eq. (3) is

2

sincossincos

2sinh'

ljlljlZ

eeZljZZ

C

ljlj

CC

ljZZ C sin (25)Likewise, we can derive, in lossless case:

2/

)2/tan('

l

lCjY (26)

Here, C is total line capacitance. Eq. (25),

(26) are same as (4.46), (4.45) in text, p.115, and we see that Z and Y are pure

reactances and susceptances, respectively,

i.e., Z=jX and Y=jB.

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17/26

Conjugating eq. (25) and (26), we get:

ljZZ C sin* (27)

2/

)2/tan('*

l

lCjY (28)

Recall eq. (24):

( )*2

1*

21

*

2

1

122/

12

YVZ

eVV

Z

VS

j

+

=

(24)

Substitution of eqs. (27), (28) into theexpression for S12, eq. (24), we get:

l

lVCj

ljZ

eVV

ljZ

VS

C

j

C

)2/tan(

sinsin

2

1

21

2

1

12

12

(29)

Bringing the j of the first term into the

numerator and canceling the two negativesigns of the second term results in:

l

lVCj

ljZ

eVV

lZ

VjS

C

j

C

)2/tan(

sinsin

2

1

21

2

1

12

12

(30)

8.0 Real power

The first and last terms of eq. (30) are purelyimaginary and so do not affect real power.

Lets look more closely at the second term:

17


18/26

lZ

jVV

lZjVV

lZ

eVV

lZ

eeVV

ljZ

eVV

C

C

C

j

C

jj

C

j

sin

cossin

sin)90sin()90cos(

sinsinsin

121221

121221

)90(

21

90

2121121212

(31)

Of this term, we see that the real part is the

sin12 term. Since this is the only real part of

eq. (30), it must be true that:

lZ

VVP

C sin

sin 122112 (32)

It is generally the case that voltage

magnitudes at either end of a line are not

very different (and when the line is losslessand terminated in ZC, they are

exactly the same see eq. (14) above). If we

assume that |V2|=|V1|, then:

lZ

VP

C sin

sin 122

1

12 (33)

Recalling eq. (17), repeated here:

C

SILZ

VP

2

1 (17)

18


19/26

we see that eq. (33) can be re-written as:

lPP SIL

sin

sin 1212 (34)

Although approximate, eq. (34) is very

useful for getting a back-of-the-envelope

sense of what voltages are required in order

to accommodate a given power transfer

level, as observed in problem 4.21 of the

homework assignment. If you take aposition as a transmission planner, you

might find this expression handy.

Problem 4.21:

Transient instability problems can occur

when the angular separation between

voltage phasors at either end of a

transmission line, denoted by 12=1-2becomes large.

So when does 12 become large?

19


20/26

Note that in eq. (34) that PSIL and sinl are

constants for an already-constructed

transmission line.

lPP SIL

sin

sin 1212 (34)

So 12 directly determines (or is directly

determined by) P12.

As 12 increases from 0 to 90, the realpower transfer P12 gets larger. Or, more

properly, we can see that as P12 gets larger,

12 increases from 0 to 90.

So answer to question of When does 12become large? is: when P12 becomes large.

Problem 4.21 indicates that a practical limit

for transmission lines on 12 is 45 (actually,

this is probably too large, a 30 limit is

better, but we will use 45).

It also indicates that the desired transfer

level on a 300 mile-long line is 500 MW.

20


21/26

Question is: what voltage levels can we

consider?

This is a design (planning) problem.

The problem also indicates =0.002/mile.

Solving eq. (34) for PSIL yields:

MWlPPSIL 40045sin

)300002.0sin(500sinsin

12

12

This is 3-phase power. Referring to Table 1

above, we see we need voltage level of at

least 345 kV to accommodate this.

9.0 Reactive power

Consider again eq. (30), repeated here:

l

lVCj

ljZ

eVV

lZ

VjS

C

j

C

)2/tan(

sinsin

2

1

21

2

1

12

12

(30)

Now lets consider the imaginary part. Indoing so, we need to remember that the

middle term contributes a real part, but it

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22/26

also contributes an imaginary part, as

indicated by eq. (31), repeated here:

lZjVV

ljZeVV

CC

j

sin

cossin

sin12122121

12

(31)

Substituting eq. (31) into eq. (30), we have:

l

lVCj

lZ

jVV

lZ

VjS

CC

)2/tan(

sin

cossin

sin

2

1

121221

2

1

12

(35)

Now taking only the imaginary part of (35):

l

lVC

lZ

VV

lZ

VQ

CC

)2/tan(

sin

cos

sin

2

1

1221

2

1

12 (36)

Recall that the denominator of the first two

terms is just |Z| in the lossless case (see eq.(25) above). Therefore, we can write:

l

lVC

Z

VV

Z

VQ

)2/tan(cos

'

2

1

1221

2

1

12 (37)

Now lets consider eq. (37) for the short line

(and lossless) model. In this case,

ZZ, and for lossless, |Z|=X C0.

So eq. (37) becomes:

22


23/26

X

VV

X

VQ

1221

2

1

12

cos(38)

This should be familiar expression to youfrom EE 303.

10.0 Voltage instability (Section 4.7)

We can also write the real power equation

(32) for the short line model, as we have just

done for the reactive power equation, using

Z=Z and |Z|=X, to get:

X

VV

lZ

VVP

C

12211221

12

sin

sin

sin(39)

Section 4.7 applies eqs. (38) and (39) toconsider how the receiving end voltage

magnitude varies with load.

Assuming the (normalized) sending-end

voltage is 1.0, the text derives:

2

)2(1122

DDD PPPV (40)

where

23


24/26

|V2| is the normalized voltage at the

receiving end,

PD is the receiving end real power load =tan, where pf=cos (note here is

different from our previous use of it).

Equation (40) can be used to plot receiving

end voltage |V2| as a function of receiving

end demand PD.

Fig. 4

24


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Fig. 4 is similar to Fig. E4.9 in the text. This

figure is a very basic figure for transmission

planners and operators, who constantlyworry about the implications of this figure.

Why?

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1[[] R. Burington, Handbook of Mathematical Tables and Formulas, 5th

Edition, McGraw-Hill.2[[] O. Elgerd, Electric Energy Systems Theory: An Introduction, 2nd edition,

McGraw-Hill, 1982.3[[] J. Glover and M. Sarma, Power System Analysis and Design, PWS

Publishers, 1987.4[[] Transmission Line Reference Book: 345 kV and Above, Electric Power

Research Institute, 2nd Edition, 1987.

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